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Functions of local operators

  1. Jan 5, 2015 #1
    Is it okay to define a local operator as an operator whose matrix elements in position space is a finite sum of delta functions and derivatives of delta functions with constant coefficients?

    Suppose your operator is M, and the matrix element between two position states is <x|M|y>=M(x,y).

    It seems that, at least formally, any function f(x,y) can be written as an infinite sum of deltas and derivative of deltas, with constant coefficients. So for example, the Green's function which is [itex]\langle x|(\partial^2-m^2)^{-1} |y\rangle=\text{BesselFunction}(x-y)[/itex] can be written as an infinite sum of deltas and derivatives of deltas, but not a finite sum, so it's not local.

    But then what about functions of the momentum operator, such as [itex]\log[P] [/itex] or [itex]e^{iP} [/itex]? Are these local?

    Is those functions aren't local, is it safe to say that only polynomials in the momentum operator can be local?

    Also, the position operator squared has this matrix element: <x|X2|y>=δ(x-y)x2. If we wanted to expand this matrix element as a sum of delta functions and derivatives with constant coefficients, it would require an infinite amount. However, if we let the coefficients depend on x, then it is a finite amount.

    So should the definition of a local operator be modified to as any operator whose matrix elements in position space is a finite sum of delta functions and derivatives of delta functions with coefficients depending only on position?
     
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  3. Jan 6, 2015 #2

    vanhees71

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    Where did you get this terminology from? Usually "local operators" are defined in relativistic quantum field theory, using the fundamental field operators as building blocks. In a local QFT, the fundamental field operators are assumed to transform as the corresponding classical fields under proper orthochronous Poincare transformations, which are realized by unitary transformations. For a translation with four-vector ##a^{\mu}## you have, e.g., for a massive vector field
    $$\hat{U}_T(a) \hat{A}^{\mu}(x) \hat{U}_T^{\dagger}(a)=\hat{A}^{\mu}(x-a)$$
    and for boosts or rotations
    $$\hat{U}_L(\Lambda) \hat{A}^{\mu}(x) \hat{U}_T^{\dagger}(\Lambda) = {\Lambda^{\mu}}_{\nu} \hat{A}^{\nu}(\Lambda^{-1} x),$$
    where ##\Lambda \in \mathrm{SO}(1,3)^{\uparrow}##.

    Such unitary representations of the proper orthochronous Poincare group are called "local realizations". A local operator is then defined as a (formal) polynom of local field operators and its space-time derivatives at the same space-time point. It's only formal, because due to the equal-time canonical commutation or anticommutation relations the field operators are operator-valued distributions that cannot be properly multiplied at the same space-time point in a strict mathematical sense.

    That's why the infinities occur in loop integrals of perturbation theory, which are cured by renormalization. There's an alternative treatment, the Epstein-Glaser approach, where these artifacts are avoided by, roughly speaking, smearing out the local field-operator products over a finite space-time volume. For a good introduction into the latter approach, see

    G. Scharf, Finite Quantum Electrodynamics, Springer
     
  4. Jan 6, 2015 #3
    Thanks. If I have a correlation function, isn't it usually infinity even if all fields in the correlator are at different spacetime points? The exception seems to be a free field theory where everything is finite: therfore it seems infinitys come from structure of hamiltonian, and not the structure of fields.

    Also, if fields are distributions, does this mean if you integrate correlation functions over space, all infinitys vanish? This does not seem to be true, because correlation functions in momentum space are infinite and also have momentum conserving delta functions, but momentum correlators are just the spacetime integrals of position ones.
     
  5. Jan 6, 2015 #4

    vanhees71

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    For correlation functions of free fields, you get well-defined functions, like the Feynman propagator in the vacuum
    $$\mathrm{i} G(x)=\langle 0|\mathrm{T}_c \hat{\phi}(x) \hat{\phi}(0)|0 \rangle.$$
    It's given as
    $$G(x)=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 x}{(2 \pi)^4} \frac{1}{p^2-m^2+\mathrm{i} 0^+} \exp(-\mathrm{i} p \cdot x),$$
    where the Minkowski product is used in the west-coast convention, i.e., ##p \cdot x=p_{\mu} x^{\mu} = p_0 x_0-\vec{p} \cdot \vec{x}##.
     
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